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IAL 2022 June Q2

A Level / Edexcel / P3

IAL 2022 June Paper · Question 2

题目

Problem

The functions ff and gg are defined by

f(x)=5−x3x+2x∈R, x≠−23f(x)=\frac{5-x}{3x+2}\qquad x\in\mathbb{R},\ x\ne -\frac23 g(x)=2x−7x∈Rg(x)=2x-7\qquad x\in\mathbb{R}

(a) Find the value of fg(5)fg(5)

(2)

(b) Find f−1f^{-1}

(3)

(c) Solve the equation

f(1a)=g(a+3)f\left(\frac1a\right)=g(a+3)
(4)
题目中文翻译

函数 ff 和 gg 定义为

f(x)=5−x3x+2x∈R, x≠−23f(x)=\frac{5-x}{3x+2}\qquad x\in\mathbb{R},\ x\ne -\frac23 g(x)=2x−7x∈Rg(x)=2x-7\qquad x\in\mathbb{R}

(a) 求 fg(5)fg(5) 的值。

(b) 求 f−1f^{-1}。

(c) 解方程

f(1a)=g(a+3)f\left(\frac1a\right)=g(a+3)

解答

(a)

First find

g(5)=2(5)−7=3g(5)=2(5)-7=3

Therefore

fg(5)=f(3)fg(5)=f(3)

Now

f(3)=5−33(3)+2=211f(3)=\frac{5-3}{3(3)+2} =\frac{2}{11}

So

fg(5)=211\boxed{fg(5)=\frac{2}{11}}

(b)

Let

y=5−x3x+2y=\frac{5-x}{3x+2}

Then

y(3x+2)=5−xy(3x+2)=5-x

So

3xy+2y=5−x3xy+2y=5-x

Collect the xx terms:

3xy+x=5−2y3xy+x=5-2y

Thus

x(3y+1)=5−2yx(3y+1)=5-2y

and hence

x=5−2y3y+1x=\frac{5-2y}{3y+1}

Therefore

f−1(x)=5−2x3x+1f^{-1}(x)=\frac{5-2x}{3x+1}

The denominator cannot be zero, so

x≠−13x\ne -\frac13

Hence

f−1(x)=5−2x3x+1,x≠−13\boxed{f^{-1}(x)=\frac{5-2x}{3x+1},\qquad x\ne -\frac13}

(c)

We need to solve

f(1a)=g(a+3)f\left(\frac1a\right)=g(a+3)

First,

f(1a)=5−1a3⋅1a+2f\left(\frac1a\right) =\frac{5-\frac1a}{3\cdot\frac1a+2}

Multiply numerator and denominator by aa:

f(1a)=5a−13+2af\left(\frac1a\right) =\frac{5a-1}{3+2a}

Also,

g(a+3)=2(a+3)−7=2a−1g(a+3)=2(a+3)-7=2a-1

So

5a−12a+3=2a−1\frac{5a-1}{2a+3}=2a-1

Multiply by 2a+32a+3:

5a−1=(2a−1)(2a+3)5a-1=(2a-1)(2a+3)

Expand the right hand side:

5a−1=4a2+4a−35a-1=4a^2+4a-3

Bring all terms to one side:

4a2−a−2=04a^2-a-2=0

Using the quadratic formula,

a=1±(−1)2−4(4)(−2)2(4)a=\frac{1\pm\sqrt{(-1)^2-4(4)(-2)}}{2(4)}

Therefore

a=1±338a=\frac{1\pm\sqrt{33}}{8}

So

a=1+338, 1−338\boxed{a=\frac{1+\sqrt{33}}{8},\ \frac{1-\sqrt{33}}{8}}